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2 years ago

a crane lifts a 35000 N steel girder a distance of 25 m in 45 s. How much power did the crane require to lift the girder

1 answer:

7 0

Hello.

The formula for Power is Work divided by Time; however, we do not have our value for Work - yet.
To find for the Work inputted, we need to use its formula: Force * Distance.

Let's multiply our Force by our Distance. Remember that our Force is always measured in Newtons (N), and our Distance is measured by Meters (M).
35,000 * 25 = 875,000 J (Unit for Work is "J" or "Joules")

Now that we have the value for Work, let's apply it to our Power formula.
P = 875,000 / 45; 19,444.44~

The Power required to lift the girder is 1944.44~ W (Unit for Power is "W" or "Watts").

I hope this helps!

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Which statement would be a valid argument in favor of using nuclear power? There are few negative impacts from mining the fuel.

fgiga [73]

Okay so, lets use the process of elimination here.

A)There are few negative impacts from mining the fuel.
B)Reactors are safe from natural disasters.
C)There are little to no waste products from fission.
D)Nuclear power does not contribute greenhouse gases.

First off, we know B cannot be correct, seeing as how reactors are fragile and are damaged easily by Japan's earthquakes. So we can eliminate B from the choices. We then can eliminate C, since fission creates high levels of nuclear waste, so that leaves us with just A, and D. We can then eliminate A since uranium is radioactive, there is always a chance for negative effects.

So, the correct answer is D

6 0

2 years ago

Read 2 more answers

3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-di

irinina [24]

Answer:

The initial velocity of the water from the tank is 5.42 m/s

Explanation:

By applying Bernoulli equation between point 1 and 2

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L$

At the point 1

P₁=0 ( Gauge pressure)

V₁= 0 m/s

Z₁=3 m

At point 2

P₂=0 ( Gauge pressure)

Z₂= 0 m/s

$h_L=1.5\ m$

Now by putting the values

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L$

$Z_1-h_L=\dfrac{V_2^2}{2g}$

$3-1.5=\dfrac{V_2^2}{2\times 9.81}$

$V_2=\sqrt{2\times 1.5\times 9.81}\ m/s$

V₂= 5.42 m/s

The initial velocity of the water from the tank is 5.42 m/s

3 0

2 years ago

The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo

irina1246 [14]

Answer:

$p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

Explanation:

first write the newtons second law:

F $_{s}$ =δma $_{s}$

Applying bernoulli,s equation as follows:

∑ $δp+\frac{1}{2} ρδV^{2} +δγz=0\\$

Where, $δp$ is the pressure change across the streamline and $V$ is the fluid particle velocity

substitute $ρg$ for {tex]γ[/tex] and $g_{0}-cz$ for $g$

$dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0$

integrating the above equation using limits 1 and 2.

$\int\limits^2_1 \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

there the bernoulli equation for this flow is $p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

note: $ρ$ =density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

4 0

2 years ago

A balloon is at a height of 81m and is ascending upwards with a velocity of 12m/s. A body of 2kg weight is dropped from it. If g

kap26 [50]

I know you are Indian by your question, HC Verma class 9 or 11 !!

if you got any problem, comment !!

7 0

2 years ago

Tom and his little sister are enjoying an afternoon at the ice rink. they playfully place their hands together and push against

Westkost [7]

Newton's third law says:
"For every action, there is an equal and opposite reaction. ".

So, the force that Tom does on the sister is equal to force the sister applies on Tom:
 $F_t = F_s$
where the label "t" means "on Tom", while the label "s" means "on the sister".

From Newton's second law, we also know
 $F=ma$
where m is the mass and a the acceleration. so we can rewrite the first equation as
 $m_t a_t = m_s a_s$
And find Tom's acceleration:
 $a_t = \frac{m_s}{m_t} a_s = \frac{15 kg}{61 kg} (2.1 m/s^2) =0.52 m/s^2$

5 0

2 years ago